Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $a = \dfrac{q + 4}{9q + 18} \div \dfrac{-3q - 12}{q^2 + 6q + 8} $
Dividing by an expression is the same as multiplying by its inverse. $a = \dfrac{q + 4}{9q + 18} \times \dfrac{q^2 + 6q + 8}{-3q - 12} $ First factor the quadratic. $a = \dfrac{q + 4}{9q + 18} \times \dfrac{(q + 2)(q + 4)}{-3q - 12} $ Then factor out any other terms. $a = \dfrac{q + 4}{9(q + 2)} \times \dfrac{(q + 2)(q + 4)}{-3(q + 4)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac{ (q + 4) \times (q + 2)(q + 4) } { 9(q + 2) \times -3(q + 4) } $ $a = \dfrac{ (q + 4)(q + 2)(q + 4)}{ -27(q + 2)(q + 4)} $ Notice that $(q + 4)$ and $(q + 2)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac{ \cancel{(q + 4)}(q + 2)(q + 4)}{ -27\cancel{(q + 2)}(q + 4)} $ We are dividing by $q + 2$ , so $q + 2 \neq 0$ Therefore, $q \neq -2$ $a = \dfrac{ \cancel{(q + 4)}\cancel{(q + 2)}(q + 4)}{ -27\cancel{(q + 2)}\cancel{(q + 4)}} $ We are dividing by $q + 4$ , so $q + 4 \neq 0$ Therefore, $q \neq -4$ $a = \dfrac{q + 4}{-27} $ $a = \dfrac{-(q + 4)}{27} ; \space q \neq -2 ; \space q \neq -4 $